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3 years ago

Compound formula: MgCl2

Chemistry

2 answers:

Tasya [4]3 years ago

8 0

Answer:

Element K

# of Atoms: 1

Element N

# of Atoms: 1

Element O

# of Atoms: 3

Gram formula weight (g): 101.11

Explanation:

umka2103 [35]3 years ago

7 0

Answer:

please where are the questions and also make your questions clear.

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nataly862011 [7]

Answer:

For A: The mass of aluminium required will be 183 g

For B: The mass of alumina produced will be $6.63\times 10^6g$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For a:

Given mass of $NH_4ClO_4$ = 1.325 kg = 1325 g (Conversion factor: 1 kg = 1000 g)

Molar mass of $NH_4ClO_4$ = 117.50 g/mol

Putting values in equation 1, we get:

$\text{Moles of }NH_4ClO_4=\frac{1325g}{117.50g/mol}=11.28mol$

For the given chemical reaction:

$6NH_4ClO_4(s)+10Al(s)\rightarrow 5Al_2O_3(s)+3N_2(g)+6HCl(g)+9H_2O(g)$

By stoichiometry of the reaction:

6 moles of $NH_4ClO_4$ reacts with 10 moles of aluminium

So, 11.28 moles of $NH_4ClO_4$ will react with = $\frac{10}{6}\times 11.28=6.77mol$ of aluminium

Now, calculating the mass of aluminium by using equation 1, we get:

Molar mass of aluminium = 27.00 g/mol

Moles of aluminium = 6.77 moles

Putting values in equation 1, we get:

$6.77mol=\frac{\text{Mass of aluminium}}{27.00g/mol}\\\\\text{Mass of aluminium}=(6.77mol\times 27.00g/mol)=183g$

Hence, the mass of aluminium required will be 183 g

For b:

Given mass of aluminium = $3.500\times 10^3=3.5\times 10^6g$ (Conversion factor: 1 kg = 1000 g)

Molar mass of aluminium = 27.00 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{3.5\times 10^6g}{27.00g/mol}=1.3\times 10^5mol$

For the given chemical reaction:

$6NH_4ClO_4(s)+10Al(s)\rightarrow 5Al_2O_3(s)+3N_2(g)+6HCl(g)+9H_2O(g)$

By stoichiometry of the reaction:

10 moles of aluminium produces 5 moles of alumina

So, $1.3\times 10^5mol$ of aluminium will react with = $\frac{5}{10}\times 1.3\times 10^5=6.5\times 10^4mol$ of alumina

Now, calculating the mass of alumina by using equation 1, we get:

Molar mass of alumina = 101.96 g/mol

Moles of alumina = $6.5\times 10^4mol$

Putting values in equation 1, we get:

$6.5\times 10^4mol=\frac{\text{Mass of alumina}}{101.96g/mol}\\\\\text{Mass of alumina}=(6.5\times 10^4mol\times 101.96g/mol)=6.63\times 10^6g$

Hence, the mass of alumina produced will be $6.63\times 10^6g$

6 0

3 years ago

If I plant a green bean seed in the desert, will it grow thick, barrel-shaped stems full of water like a cactus? Why or why not?

Ket [755]

Answer:

No, although the bean crop can grow perfectly in full sun or a little shade in very warm places. It does not mean that they will adopt the growth patterns of the cactus, especially because this plant needs constant watering (in small quantities), very constant. Beans require good drainage for their growth.

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3 years ago

Athe gas that siports burning is

kondaur [170]

The answer is E it’s killing all oxygen that is around you

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4 years ago

METHANOL + ETHANOIC ACID

Sergio [31]

Answer:

if you’re looking for a balanced equation it would be:

Methanol +ethanoic acid ==> methyl ethanoate + water ( in the presence of concentrated sulfuric acid )

CH3OH + CH3COOH==> CH3COOCH3 + H2O

~~~~~~

But, if you were looking for what it would be called it would be:

ethanoic acid with methanol will produce methyl ethanoate.

~~~~~~

And the reaction between Methanol and Ethanoic acid is known as esterification.

~~~~~~

I wasn’t sure of your question so I answered the best way that I could. I hope this helped!

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3 years ago

Calculate the molar concentration of the Cl⁻ ions in 0.87 M MgCl2(aq).

Fed [463]

The compound MgCl2(aq) is ionic which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of MgCl2 that dissolves.
MgCl2(s) --> Mg+(aq) + 2 Cl⁻(aq)
[Cl⁻] = 0.87 mol MgCl2/1L × 2 mol Cl⁻ / 1 mol MgCl2 = 1.7 M
The answer to this question is [Cl⁻] = 1.7 M

8 0

3 years ago